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I know the first few terms are:

$$ \log\cosh(x) = \frac{x^2}{2} - \frac{x^4}{12} + \frac{x^6}{45} + \mathcal{O}(x^8) $$

Is there a closed-form expression for the $n$'th coefficient of this expansion?

a06e
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    From numerator (A046990) and denominator (A046991), the coefficients appear to be something like $$\frac{2^{2n-1} (2^{2n}-1) B(2n)}{(2n)! n,}$$ where $B(n)$ are the Bernoulli numbers (as explained in the FORMULA section). However, those sequences are apparently also the coefficients for the series of $\log \cos x$ (or at least, the absolute values are). I believe I've corrected this by changing $|B(2n)|\mapsto B(2n)$, but there might be problems with the indexing or signs of these terms. – Graviton Apr 28 '23 at 08:25
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    @Graviton i think the reason for that is the identity $$\cos(ix)=\cosh(x)$$ and the fact that the powers of $x$ are all even. – Zima Apr 28 '23 at 08:56
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    @Zima Ah! That's a nice catch. – Graviton Apr 28 '23 at 08:59

1 Answers1

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The absolute value of your numerators and denominators appear to be A046990 and A046991 respectively. There, they give the formula $$\frac{\text{A046990}(n)}{\text{A046991}(n)} = \frac{2^{2n-1}\left(2^{2n}-1\right)|B_{2n}|}{(2n)!n},$$

where $B_{2n}$ are the Bernoulli numbers,

$$B_{2n}=\frac{(-1)^{n-1}2(2n)!}{(2\pi)^{2n}}\zeta(2n).$$

However, these are interestingly not the coefficients for $\log\cosh x$, but rather for $\log \cos x$ (Edit: thanks for @Zima for pointing out that this is almost certainly due to $\cos ix=\cosh x$). Hence, since your coefficients are signed, we may guess that $|B_{2n}|\mapsto B_{2n}$ recovers the signs. Visual inspection seems to agree, however I lack proof.

In any case, I have strong evidence for the following closed form,

$$\log\cosh x = \sum_{n=1}^\infty \frac{2^{2n-1}\left(2^{2n}-1\right)B_{2n}}{(2n)!n}x^{2n}.$$

If we "simplify" by plugging in the above formula for $B_{2n}$,

$$\log\cosh x = \sum_{n=1}^\infty \frac{(-1)^{n-1}\left(2^{2n}-1\right)\zeta(2n)}{n\pi^{2n}}x^{2n}.$$

It may be possible to clean this up more by the zeta functional equation but I'll leave it as-is.

Graviton
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    The result seems correct. I believe you can prove it just by integrating the well known series for $\tan(x)$, and this also explains the appearance of the Bernoulli numbers. – Zima Apr 28 '23 at 10:02